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Thermochemistry Chapter 5 What is Energy? Sources of Energy 5.1 The Nature of Energy An object?s internal energy, Einternal, is the sum of its potential energy and its kinetic energy: Einteral = PE + KE Kinetic Energy Kinetic Energy: Energy of Motion KE = Ek= 1/2 m?2 ?T Kinetic Energy depends upon the mass and velocity of the object that is moving. Potential Energy Potential Energy: Energy of Position stored energy due to forces of attraction or repulsion between objects such as: gravitational forces magnetic forces electrostatic forces chemical energy Potential and Kinetic Energy can be inter-converted System and Surroundings To describe energy changes, we must define the system and the surroundings. In a closed system, energy can be exchanged, but not matter. system surroundings surroundings surroundings Thermal Equilibrium 5.2 The First Law of Thermodynamics Energy is conserved Energy cannot be created or destroyed. Energy can be transferred back and forth between the system and the surroundings. Energy can be converted to different types of energy. Change in Energy (?E) E of system increases Change in Energy (?E) Example Change in Energy (?E) Energy - the capacity to do work (w) or transfer heat (q) ?E = Change in Energy (Ef - Ei) ?E = q + w Work (w) - Energy used to cause an object to move against force (w = F × d) Heat (q) - Energy transferred from a hotter object to a colder object. Kinetic and Potential Energy Suppose that the ball is the system. Work on the system by surroundings increases the PE of the ball. As the ball falls, PE decreases as KE increases. When the ball hits the ground, KE=0 and energy is released to surroundings as heat. ?E = q + w Einteral = PE + KE Sign Conventions for Heat and Work Work (w) Work can be calculated as PV work. When pressure is constant: w = ?P?V The Sign of w The value of w is positive when work is done on the system. Example: A gas is compressed The value of w is negative when work is done by the system. Example: A gas expands The Sign of q The value of q is positive when heat is absorbed by the system (ENDOTHERMIC) Examples: Evaporation; The value of q is negative when heat is given off by the system (EXOTHERMIC) - Examples: Acid-base reactions; Explosions; State Functions A state function is a property of a system that depends upon conditions such as temperature, pressure, etc, but does NOT depend upon how the sample got to its present condition. Internal Energy, E, is a state function. Not q and w The internal energy (E) of 50 g of water at 25°C does not depend on its history The change in internal energy, ?E, is also State Function 50 g Ice at -5 ?C 50 g water vapor at 100?C 50 g water at 25?C E 5.3 Enthalpy Change (?H) (A state function) When a process occurs at constant pressure the heat gained or lost by the system, q, is equal to the enthalpy change of the system: ?H = qp Most reactions occur at constant pressure open beakers allow changes in volume without the build up of pressure. When does ?H = ?E? The enthalpy change equals the change in internal energy when work is negligible. In most cases, even when gases are formed, the calculated amount of work is very small relative to ?H. Therefore ?E ? ?H. Energy Units The SI unit for energy is the Joule (J) 1 Joule (J) = 1 N?m = 1 kg ?m2/s2 1 calorie (cal) is the amount of energy required to increase the temperature of 1.0 g of water by 1?C. 1 cal = 4.184 J Nutritional calories: 1 Cal = 1 kcal = 1000 cal 5.5 Calorimetry Calorimetry is an experimental method for measuring heat flow (q or ?H) by measuring the temperature change (?T) of the system or the surroundings. How is q related to ?T? Heat Capacity and Specific Heat Heat Capacity (C) The amount of heat required to increase the temperature of an object by 1 ?C. Units: J/?C Molar Heat Capacity (Cm) The heat capacity for 1 mol of substance Units: J/mol?C Specific Heat The heat capacity for 1 gram of substance Units: J/g?C Heat Capacity and Specific Heat Molar heat capacity and specific heat values are intrinsic properties of substances. Heat = (Heat Capacity)( ?T) Heat = (Molar Heat Capacity)(#mol)(?T) Heat =(Specific Heat)(mass)(?T) Specific Heat Values (T 5.2) Heat =(Specific Heat)(mass)(?T) Specific Heat (J/g?C) Ethanol(l) 2.43 H2O(l) 4.184 Al(s) 0.90 Fe(s) 0.45 Which substance requires the most amount of heat to increase the temperature of 1 gram by 1?C? Specific Heat Values (T 5.2) Heat =(Specific Heat)(mass)(?T) Specific Heat (J/g?C) Ethanol(l) 2.43 H2O(l) 4.184 Al(s) 0.90 Fe(s) 0.45 If the same amount of heat is added to equal masses of these substances, which would increase in temperature by the greatest amount? First Law of Thermodynamics Energy is conserved. The amount of heat lost by a system is equal to the amount of heat gained by the surroundings. The amount of heat gained by a system is equal to the amount of heat lost by the surroundings. qsystem = ? qsurroundings Heat Calculations When a hot rock is added to 50.0 g of water, the water increases in temperature from 25 to 35?C. How much heat is given off by the rock? Calculating q for Reactions Often we do not know the specific heat or ?T for the system. Instead we determine qsurroundings, and then use the 1st law of thermodynamics: qsystem = ? qsurroundings qreaction = ? qsurroundings Constant-Pressure Calorimetry ?H= q When 50.0 mL of 1.0 M HCl and 50.0 mL of 1.0 M NaOH are mixed in a coffee cup calorimeter, the temperature of the resultant solution increases from 21.0?C to 27.5?C. Assume: Calorimeter does not absorb a significant amount of heat Density of solution is 1.0 g/mL Specific Heat of the solution is 4.184 J/g?C. Calculate: Enthalpy change for the Neutralization Reaction, ?Hrxn? Constant-Pressure Calorimetry ?H= q Find q of the surroundings (the reacting species is your system): qwater = (sp. heat)(mass)(?T) qwater = (4.184 J/g?C)(100 g)(27.5?21.0?C) = +2719.6 J Constant-Pressure Calorimetry ?H= q Find q of the surroundings (the reacting species are your system): qwater = (sp. heat)(mass)(?T) qwater = (4.184 J/g?C)(100 g)(27.5?21.0?C) = +2719.6 J qrxn = ?qwater = ?2719.6 J Constant-Pressure Calorimetry ?H= q Find q of the surroundings (the reacting species are your system): qwater = (sp. heat)(mass)(?T) qwater = (4.184 J/g?C)(100 g)(27.5?21.0?C) = +2719.6 J qrxn = ?qwater = ?2719.6 J = -2720 J = -2.72 kJ Constant-Pressure Calorimetry ?H= q qrxn = ?2.72 kJ What is the ?Hrxn in kJ/mol? How many mol of HCl reacted? Constant-Pressure Calorimetry ?H= q qrxn = ?2.72 kJ What is the ?Hrxn in kJ/mol? How many mol of HCl reacted? (1.0 mol/L)(0.050 L) = 0.050 mol Constant-Pressure Calorimetry ?H= q qrxn = ? 2.72 kJ What is the ?Hrxn in kJ/mol? How many mol of HCl reacted? (1.0 mol/L)(0.050 L) = 0.050 mol ?Hrxn = ? 2.72 kJ/ 0.050 mol = -54 kJ/mol Enthalpy Changes for Phase Changes ?H?vap Liquid phase to Gas Phase ?H?fus Solid Phase to Liquid Phase ?H?sub Solid Phase to Gas Phase Should these be positive values or negative values? Group Work Identify each of the following as an endothermic or exothermic process: H2O(l) ? H2O(g) H2O(l) ? H2O(s) Br2(g) ? Br2(l) I2(s) ? I2(g) Enthalpy Changes for Phase Changes Identify each of the following as an endothermic or exothermic process: H2O(l) ? H2O(g) Endothermic H2O(l) ? H2O(s) Exothermic Br2(g) ? Br2(l) Exothermic I2(s) ? I2(g) Endothermic 5.4 Enthalpy of Reaction (?Hrxn) ?Heat of Reaction? The enthalpy change, ?H, for a reaction is a measure of the heat gained or lost during the reaction. ?H is also a good estimate of the internal energy change for a system as it undergoes a chemical change. 5.4 Enthalpy of Reaction (?Hrxn) ?Heat of Reaction? ?Hrxn = ?H(products) ? ?H(reactants) Thermochemical equation: (balloon) 2H2(g) + O2(g) ? 2H2O(g) ?H = -483.6 kJ This is the enthalpy change for the reaction when 2 mol H2 react completely with 1 mol O2. Enthalpy of Reaction (?Hrxn) ?Heat of Reaction? ?Hrxn = ?H(products) ? ?H(reactants) Thermochemical equation: 2H2(g) + O2(g) ? 2H2O(g) ?H = -483.6 kJ 2H2(g) + O2(g) reactants 2H2O(g) product ?H Enthalpy of Reaction (?Hrxn) ?Heat of Reaction? ?Hrxn = ?H(products) ? ?H(reactants) Thermochemical equation: 2H2(g) + O2(g) ? 2H2O(g) ?H = -483.6 kJ What is the enthalpy change when 1 mol H2 reacts completely? Enthalpy of Reaction (?Hrxn) ?Heat of Reaction? ?Hrxn = ?H(products) ? ?H(reactants) Thermochemical equation: 2H2(g) + O2(g) ? 2H2O(g) ?H = -483.6 kJ What is the enthalpy change when 10 mol H2 reacts completely? Guidelines for ?Hrxn 1. Enthalpy is an extensive property. ?H is dependent upon amounts. 2. Reverse take the opposite sign. 2H2O(g) ? 2H2(g) + O2(g) ?H = +483.6 kJ 3. ?H depends upon the physical state of reactants and products (g, l, s, aq). 2H2O(l) ? 2H2(g) + O2(g) ?H = +571.7 kJ 5.6 Hess?s Law Hess?s Law: If a reaction is carried out in a series of steps, ?H for the reaction will be equal to the sum of the enthalpy changes for the individual steps. CH4(g) + 2O2(g) ? CO(g) + 2H2O(l) + ½ O2(g) ?H2 = -607 kJ CO(g) + ½ O2(g) ? CO2(g) ?H3 = -283 kJ ------------------------------------------------------------------------------------ CH4(g) + 2O2(g) ? CO2(g) + 2H2O(l) ?H1 = -890 kJ Figure 5.21 Hess?s Law Example (Like Examples 5.8 and 5.9) N2(g) + 2O2(g) ? 2NO2(g) ?H1 = 66.4 kJ NO(g) + ½O2(g) ? NO2(g) ?H2 = ?57.1 kJ ------------------------------------------------------------------------ N2(g) + O2(g) ? 2NO(g) ?H3 = ? 5.7 Enthalpies of Formation ?Hf Enthalpy of formation is the ?H for the reaction that forms a compound from its constituent elements. Standard enthalpy of formation is when reactants and products are in their most stable forms (standard state). Formation Reactions: 3C(s) + 4H2(g) ? C3H8(g) C(s) + O2(g) ? CO2(g) H2(g) + 1/2O2(g) ? H2O(l) Group Work The standard enthalpy of formation, ?Hf?, of solid silver nitrate is ?124.4 kJ/mol. Write the formation reaction that corresponds to this value. Standard Enthalpies of Formation, ?H?f ?H?f - Standard Enthalpy of Formation Specifically for a formation reaction with reactants and products in their standard states ?H? - Standard Enthalpy Change -obtained from ?H?f values. Using Standard Enthalpies of Formation (?H?f ) Because ?H?rxn is a state function, we can theoretically accomplish this overall reaction by decomposing the reactants to its elements, then forming the product compounds from their elements. Elements Reactants Products Opposite sign of ?H?f values (× #moles) ?H?f values (× # moles) Using Standard Enthalpies of Formation (?H?f ) C3H8(g) + 5O2(g) ? 3CO2(g) + 4H2O(l) C, H2 and O2 C3H8 and O2 CO2 and H2O Opposite sign of ?H?f values (x #moles) ?H?f values (× # moles) Using Standard Enthalpies of Formation (?H?f ) We can theoretically accomplish this reaction by decomposing the reactants to its elements, then forming the product compounds from their elements. C3H8(g) ? 3Cgraphite + 4H2(g) ?H?f= -103.85 kJ (?) 3Cgraphite + 3O2(g) ? 3CO2(g) ?H?f= -393.5 kJ (x3) 4H2(g) + 2O2(g) ? 4H2O(l) ?H?f= -285.8 kJ (x4) C3H8(g) + 5O2(g) ? 3CO2(g) + 4H2O(l) ?H? = -2219.9 kJ General Equation for Calculating ?H?f from ?H?rxn values The use of Hess?s Law for determining ?H?rxn values from ?H?f values is summarized by the following equation: ?H?rxn = ? n ?Hf? (products) ? ? n ?Hf? (reactants) n = number if moles of each reactant and products in the thermochemical equation C3H8(g) + 5O2(g) ? 3CO2(g) + 4H2O(l) ?H? = General Equation for Calculating ?H?f from ?H?rxn values ?H?rxn = ? n ?Hf? (products) - ? n ?Hf? (reactants) ?H?rxn = 3(-393.5 kJ) + 4(-285.8 kJ) - (-103.85 kJ) = -2219.9 kJ C3H8(g) + 5O2(g) ? 3CO2(g) + 4H2O(l) ?H? = -2219.9 kJ General Equation for Calculating ?H?f from ?H?rxn values If you have access to ?H?f values for reactants and products, use: ?H?rxn = ? n ?Hf? (products) - ? n ?Hf? (reactants) (See End-of-chapter problems #71 and 72) If you know ?H? values for reactions that combine to give overall reactions, use the more general form of Hess?s Law: Manipulate the equations so that you can add them and their ?H? values (Hess?s Law, #59-64) Group Work ?H?rxn = ? n ?Hf? (products) - ? n ?Hf? (reactants) Calculate ?H?rxn for the following equation: 2SO2(g) + O2(g) ? 2SO3(g) ?H?f (SO2(g)) = ?296.9 kJ/mol ?H?f (SO3(g)) = ?395.2 kJ/mol Group Work Calculate ?H?rxn for the following equation: 2SO2(g) + O2(g) ? 2SO3(g) ?H? = 2(?395.2 kJ/mol) ?2(?296.9 kJ/mol) = ?196.6 kJ